Find the locations and values of any global extrema of f(x)=1-x^2 on the interval [2,2]

To find the locations and values of any global extrema of the function f(x) = 1 - x^2 on the interval [2, 2], we first need to determine whether there are any local extrema on this interval.

1. Check for critical points: Critical points occur where the derivative of the function is either zero or undefined. Let's find the derivative of f(x):
f'(x) = -2x

To find the critical points, we set f'(x) equal to zero and solve for x:
-2x = 0
x = 0

We have one critical point at x = 0. However, x = 0 is not in the interval [2, 2], so it is not relevant to our problem.

2. Evaluate f(x) at the endpoints: In this case, the interval is [2, 2]. Since there are no other points within the interval, we need to evaluate the function at the endpoints:
f(2) = 1 - (2)^2
= 1 - 4
= -3

The function value at both endpoints is -3.

Since there are no other critical points within the interval, we can conclude that the global extrema of f(x) = 1 - x^2 on the interval [2, 2] occur at x = 2 and the function values are both -3.

Therefore, the locations of the global extrema are x = 2, and the values of the global extrema are -3.